# What's wrong in this proof?(double limits and iterated limits)

It is true that if $$f$$ is a real valued function, and if the iterated limits $$\lim_{x\rightarrow x_0}[\lim_{y\rightarrow y_0}f(x,y)]$$ and $$\lim_{y\rightarrow y_0}[\lim_{x\rightarrow x_0}f(x,y)]$$ exist and are equal, we cannot say anything about the double limit $$\lim_{(x,y)\rightarrow (x_0,y_0)} f(x,y)$$ .

But look at this proof. Let $$\lim_{y\rightarrow y_0}f(x,y)=l_{y_0}(x)$$, $$\lim_{x\rightarrow x_0}f(x,y) = l_{x_0}(y)$$, $$\lim_{y\rightarrow y_0}[\lim_{x\rightarrow x_0}f(x,y)] = l_{x_0 y_0}$$

Let $$\epsilon$$> 0. $$\exists \ \delta _1 >0$$ st whenever $$|x-x_0|< \delta_1$$ we have $$|f(x,y)-l_{x_0}(y)|<\epsilon$$ and $$\exists$$ $$\delta _2 >0$$ st whenever $$|y-y_0|< \delta_2$$ we have $$|l_{x_0 y_0}-l_{x_0}(y)|<\epsilon$$.

Then we have $$|f(x,y)-l_{x_0 y_0}|\leq |f(x,y)-l_{x_0}(y)|+ |l_{x_0 y_0}-l_{x_0}(y)|<2\epsilon$$ for $$|x-x_0|<\delta_3$$ and $$|y-y_0|<\delta_3$$ where $$\delta_3= min(\delta_1, \delta_2)$$

We can proceed similarly with $$\lim_{x\rightarrow x_0}[\lim_{y\rightarrow y_0}f(x,y)]$$ and get the same result (although it is not necessary). Thus ,we have shown the double limit is indeed the same as the iterated limit. Where did I go wrong in the proof?

• Your definition of $l_{y_0}$ is unclear. Where did the $x$ go? The expression $\lim_{y\rightarrow y_0} f(x,y)$ assumes the limit exists for all $x$, and in general the limit would depend on $x$. So it would make more sense to say $\lim_{y\rightarrow y_0} f(x,y) = l_{y_0, x}$ (assuming the limit exists). Example: If $f(x,y)=xy$ then $\lim_{y\rightarrow y_0} f(x,y)= xy_0$. Oct 27, 2020 at 13:26
• @Michael I have made the necessary edits. Thanks for pointing that out. Oct 27, 2020 at 13:41
• As mentioned in an answer below, you really have $\delta_1(y)$ and $\delta_2(x)$. You would need to again rewrite the proof. Also, in your paragraph "Then we have ...", it is not clear what assumptions you make on $x$ and $y$. Oct 27, 2020 at 14:13
• @Michael I think I get it now. Thanks. Oct 27, 2020 at 14:19

In your proof, $$\delta_1$$ depends on $$y$$. So, it may happen that $$\delta_1$$ shrinks to $$0$$, as $$y\to y_0$$. Consider the function $$f\colon \mathbb{R}^2\to \mathbb{R}$$,

$$f(x,y)=\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ for $$x,y\neq 0$$ and $$f(0,0):=0$$.

Let $$(x_0,y_0):=(0,0)$$. Now, it is easy to see, that (in your notation) $$l_{x_0}=l_{y_0}=l_{x_0 y_0}=0$$, but $$f(\delta,\delta)=1$$ for any $$\delta>0$$, so that $$\lim_{(x,y)\to(x_0,y_0)}f(x,y)$$ does not exist (note that $$f(0,0)=0$$ by definition). Now, why does your proof don't work in this example? Well, for $$0<\varepsilon<1$$, no matter how small you choose $$\delta_1>0$$, we have for $$x$$ such that $$|x-x_0|<\delta_1$$ and $$y=x$$

$$|f(x,y)-l_{x_0}(y)|=|f(x,y)|=1>\varepsilon$$.

So in this example, if $$y$$ is of the same magnitude as $$x$$, then you cannot ensure $$|f(x,y)-l_{x_0}(y)|<\varepsilon$$ for $$|x-x_0|<\delta_1$$. You can only ensure it if you fix $$y$$, i.e., $$\delta_1=\delta_1(y)$$ depends on $$y$$ and $$\delta_1(y)\to 0$$ as $$y\to y_0$$ so that $$\delta_3(y)=\min\{\delta_1(y), \delta_2\}\to 0$$ as $$y\to y_0$$, contradicting the requirement that $$\delta_3$$ has to be strictly positive.

• But when y is in a neighborhood of y_0, \delta_1 must be non zero Oct 27, 2020 at 13:58
• @suraj : You are stating that $\delta_1(y)>0$ for all $y$. Fine. But so what? Oct 27, 2020 at 14:14
• @suraj: I editted my answer and gave an example. I hope that clarifies more why your proof does not work.
– ym94
Oct 27, 2020 at 14:35